$n_M=\dfrac{4,11}{M}$
$M+nH_2O\to M(OH)_n+ 0,5nH_2$
$\Rightarrow n_{M(OH)_n}=\dfrac{4,11}{M}$
$n_{H_2}=\dfrac{2,055n}{M}$
$m_{\text{dd spứ}}=4,11+81,45-\dfrac{2,055n}{M}.2=85,56-\dfrac{4,11n}{M}(g)$
$C\%=6\%$
$\Rightarrow \dfrac{4,11}{M}.(M+17n)=0,06(85,56-\dfrac{4,11n}{M})$
$n=2\Rightarrow M=137(Ba)$
Vậy M là bari.